a) 0.10 M NH4Cl b) 0.10 M H3BO3 c) 0.10 M Na2CO3 d) 0.10 M Na2CO3 mixed with 0.1
ID: 544341 • Letter: A
Question
a) 0.10 M NH4Cl
b) 0.10 M H3BO3
c) 0.10 M Na2CO3
d) 0.10 M Na2CO3 mixed with 0.10 M NaHCO3
Give an exmaple of how one of these pH values would be used for the evaluation of equilibrum constant.
Please show work and equations for all of the above.
Chem102 Manual-Final x Prelab Assignments Write the balanced net ionic equations (c -j only) to illustrate the reaction(s) occurring between the solute and water. Use a RICE box to calculate the theoretical pH value for each of these solutions. Question (c) has been started for you as an example. Write your answer for each solution in the box provided. You must show all of your work for full credit. A correct answer with no work and reasoning shown is worth no points, but an incorrect or partial answer with some work shown MAY be worth partial credit. Useful Equilibrium Constants HaBO HPoa) NH a) Deionized water (DI-H20) = 1.8x10- 5.4x10 = 5.6x 10. 4.3x10 Kal = 7.5x10" , Ka2=62x10". Ka3 = 4.2x10- 13 k,= 1.8x 10. b) 0.10M HCI c) 0.10M HOAc (acetic acid) O Type here to search 5:39 PM 10/19/2017Explanation / Answer
a) Kb for NH3 = 1.8*10-5; Ka = Kw/Kb = (1.0*10-14)/(1.8*10-5) = 5.5*10-10 (since Ka*Kb = Kw).
The acid dissociation equation is
NH4Cl (aq) --------> NH3 (aq) + H+ (aq)
Ka = [NH3][H+]/[NH3] = (x).(x)/(0.10 – x)
====> 5.5*10-10 = x2/(0.10 – x)
Since Ka is small, we can assume x << 0.10 M and hence,
5.5*10-10 = x2/0.10
====> x2 = 5.5*10-11
====> x = 7.416*10-6
Therefore, [H+] = 7.416*10-6 M and pH = -log [H+] = -log (7.416*10-6) = 5.1298 5.13 (ans).
b) The acid dissociation equation is
H3BO3 (aq) --------> H2BO3- (aq) + H+ (aq)
Ka = [H2BO3-][H+]/[H3BO3] = (x).(x)/(0.10 – x)
====> 5.4*10-10 = x2/(0.10 – x)
Since Ka is small, we can assume x << 0.10 M and hence,
5.4*10-10 = x2/0.10
====> x2 = 5.4*10-11
====> x = 7.348*10-6
Therefore, [H+] = 7.348*10-6 M and pH = -log [H+] = -log (7.348*10-6) = 5.1338 5.13 (ans).
c) Consider the step-wise dissociation of H2CO3.
H2CO3 (aq) --------> H+ (aq) + HCO3- (aq); Ka1 = 4.3*10-7
HCO3- (aq) --------> H+ (aq) + CO32- (aq); Ka2 = 5.6*10-11
We have CO32- (from Na2CO3) which undergoes acid-base equilibrium as
CO32- (aq) + H2O (l) -------> HCO3- (aq) + OH- (aq)
Since OH- is formed, we must find out Kb2. Kb2 = Kw/Ka2 = (1.0*10-14)/(5.6*10-11) = 1.8*10-4.
Kb = [HCO3-][OH-]/[H2CO3] = (x).(x)/(0.10 – x)
====> 1.8*10-4 = x2/(0.10 – x)
Since Kb is small, we can assume x << 0.10 M and hence,
1.8*10-4 = x2/0.10
====> x2 = 1.8*10-5
====> x = 4.2426*10-3
Therefore, [OH-] = 4.2426*10-3 M and pOH = -log [OH-] = -log (4.2426*10-3) = 2.3724; therefore, pH = 14.00 – pOH = 14.00 – 2.3724 = 11.6276 11.63 (pH + pOH = 14.00) (ans).
d) Since the mixture contains Na2CO3 and NaHCO3, we will work with Ka2. Given Ka2 = 5.6*10-11, pKa2 = -log (Ka2) = 10.25.
Use the Henderson-Hasslebach equation as
pH = pKa2 + log [Na2CO3]/[NaHCO3] = 10.25 + log (0.10 M)/(0.10 M) = 10.25 + log (1.0) = 10.25 (since log (1) = 0) (ans).
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.